Problem: $g(t) = -7t^{2}-7t$ $f(x) = -5x^{2}-4(g(x))$ $h(t) = 6t-3(f(t))$ $ g(f(-1)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = -5(-1)^{2}-4(g(-1))$ To solve for the value of $f$ , we need to solve for the value of $g(-1)$ $g(-1) = -7(-1)^{2}+(-7)(-1)$ $g(-1) = 0$ That means $f(-1) = -5(-1)^{2}+(-4)(0)$ $f(-1) = -5$ Now we know that $f(-1) = -5$ . Let's solve for $g(f(-1))$ , which is $g(-5)$ $g(-5) = -7(-5)^{2}+(-7)(-5)$ $g(-5) = -140$